Đặt \[g\left( x \right) = \frac{{f\left( x \right) - 15}}{{x - 2}}\]\[ \Rightarrow f\left( x \right) = \left( {x - 2} \right)g\left( x \right) + 15\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left[ {\left( {x - 2} \right)g\left( x \right) + 15} \right] = 15\].

Ta có: \[\mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 15}}{{\left( {{x^2} - 4} \right)\left( {\sqrt {2f\left( x \right) + 6} + 3} \right)}}\]\[ = \mathop {\lim }\limits_{x \to 2} \frac{{f\left( x \right) - 15}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {\sqrt {2f\left( x \right) + 6} + 3} \right)}}\]

\[ = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{f\left( x \right) - 15}}{{x - 2}} \cdot \frac{1}{{\left( {x + 2} \right)\left( {\sqrt {2f\left( x \right) + 6} + 3} \right)}}} \right]\]\[ = 3 \cdot \frac{1}{{4 \cdot \left( {\sqrt {2 \cdot 15 + 6} + 3} \right)}} = 3 \cdot \frac{1}{{4 \cdot 9}} = \frac{1}{{12}}\].

Đáp án: \[\frac{1}{{12}}\].